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From: James Waldby on 15 Jul 2010 20:24
On Thu, 15 Jul 2010 15:38:27 -0700, Ludovicus wrote:
> On 15 jul, 08:30, Harald Helfgott wrote:
>> > Your Polymath's citation is a superseded question. The AKS algorithm
>> > determine primality in (log n)^k ophenerations, [Much less than
>> > n^(1/2)]

>> Not so. The fact that the AKS algorithm determines primality in (log
>> n)^k operations doesn't mean
>> that the AKS algorithm can *find* a prime between N and 2N in (log n)^k
>> operations.

> Why not?
> First you form the number N0 = 6*floor[N/6],then N1 = 6*N0 + 1 Then
> produce the sequence U(n)= U(n)+ 4 ---> U(n+1)= U(n)+ 2 beginig with
> U(0) = N1. Before U(n) attains N1 + (Log N)^2 that is in [Log N)^2]/3
> steps ,you will have a prime number ready for applying AKS.

Besides the problem that Gerry Myerson mentioned, and besides
having a misspelled beginning, your method as stated might
return a number less than N. For example, if N=99, you have
N0=96, N1=97, and U(0)=97, which is prime but less than N.
(How serious this error is isn't clear from the context.)

--
jiw
From: Ludovicus on 16 Jul 2010 12:20
On 15 jul, 19:32, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email>
wrote:

> You seem to be suggesting that there's always a prime
> between n and n + (log n)^2. A lot of people would be
> interested in seeing a proof of such an allegation.
> Gerry Myerson (ge...(a)maths.mq.edi.ai) (i -> u for email)

Yes. I'm sorry. I followed an Archimedian path.
Ludovicus
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